CHAPTER 4
SELECTIVE HARMONIC ELIMINATED PULSE WIDTH
MODULATION (SHE PWM)
The Selective Harmonic Eliminated PWM (SHE PWM) technique is currently
applied in conventional three-level inverter circuits. The concept of the SHE PWM
technique will be presented in this chapter. It needed to be compared to the optimized
harmonic stepped-waveform technique in several aspects. Mainly, the harmonic
components and the harmonic characteristics will be focused.
4.1 Introduction
The SHE PWM technique is currently used to synthesize an output waveform of
both a half-bridge and a full-bridge inverter. In this thesis, a three-level SHE PWM
generated by a full-bridge inverter is considered. A full-bridge or H-bridge voltage source
inverter, which comprises four switches and one dc source, is depicted in Fig. 4.1.
Three states of an output waveform such as positive, negative, and zero, can be
obtained. Fig 4.2 shows a generalized three-level SHE PWM waveform, which is
synthesized by using the inverter circuit shown in Fig 4.1. The output waveform is
chopped N times per quarter. Each switch is, therefore, switched N times per cycle to
generate such a waveform.
62
S1
S2
A
Vout
E
B
S4
S3
Figure 4.1 A full-bridge voltage source inverter.
Vout
2N chops per half-cycle
+E
ωt
0
-E
α1 α2 … αN
…
αN+1 αN+2 α2N
Figure 4.2 Generalized three-level SHE PWM waveform.
4.2 The SHE PWM Waveform
The H-bridge inverter as discussed above has three states of operation. Consider
the generalized three-level SHE PWM waveform shown in Fig. 4.2. Let N be the number
of switching angles per quarter-cycle. The output waveform is assumed to be odd quarterwave symmetry, whose amplitude equals E.
63
4.2.2 The Fourier Series of the SHE PWM Waveform
Because of odd quarter-wave symmetry, the dc component and the even
harmonics are equal to zero. Patel and Hoft [5] presented the Fourier series of the threelevel SHE PWM as follows:
v out (ωt ) =
∞
∑a
sin( nωt )
(4.1)
, for odd n
(4.2)
n
n =1
where
an =
4E
nπ
N
∑ (−1)
k +1
cos( nα k )
k =1
N is the number of the switching angles per quarter.
α k is the switching angles, which must satisfy the following condition:
α1 < α 2 < K < α N <
π
2
(4.3)
E is the amplitude of the dc source.
and n is the harmonic order.
4.2.3 Apply the Newton’s method to solve SHE PWM switching angles
From (4.2), the nonlinear equation system of SHE PWM waveform can be written
as follows:
cos(α 1 ) − cos(α 2 ) + ... ± cos(α N ) =
π
M
4
cos(3α 1 ) − cos(3α 2 ) + ... ± cos(3α N ) =
64
3π
h3
4E
(4.4a)
(4.4b)
cos(5α 1 ) − cos(5α 2 ) + ... ± cos(5α N ) =
5π
h5
4E
(4.4c)
M
cos( Nα 1 ) − cos( Nα 2 ) + ... ± cos( Nα N ) =
Nπ
hn
4E
(4.4n)
where
M is the modulation index, and M =
h1
.
E
From equations (4.4a) to (4.4b), the cosine terms of α N are negative with even N
and positive with odd N. To control the amplitude of the fundamental component, the
modulation index in (4.4a), M, is given. According to the above nonlinear system, N-1
surplus harmonic can be eliminated from the output waveform by setting equations (4.4b)
to (4.4n) to zero. Basically, the lowest odd harmonic components need to be eliminated
from a single-phase system. In a three-phase system, the lowest non-triplen harmonic
components are eliminated. Generally, all triplen-harmonics in line-to-line voltage will be
eliminated by 120 electrical degree phase shift characteristic.
Example 4.1
A three-level five-switching angle SHE PWM waveform, which is shown in Fig.
4.3, will be used as an example to explain how to apply the Newton-Raphson method to
solve the SHE PWM switching angles. In this example, a single-phase system is
assumed, and modulation index of output voltage, M, is 0.85.
65
Vout
+E
0
α1 α2 ... α5
2π
π/2
ωt
-E
Figure 4.3 Output voltage waveform of a three-level five-angle SHE PWM.
From the output waveform shown in Fig. 4.3, five unknowns, α1 to α5, need to be
solved. Because of a single-phase system, the lowest four odd harmonics, i.e., the 3rd, 5th,
7th, and 9th will be eliminated. From equations (4.4a) to (4.4n), five nonlinear equations
will, therefore, be set up as follows:
π
4
(4.5a)
cos(3α 1 ) − cos(3α 2 ) + cos(3α 3 ) − cos(3α 4 ) + cos(3α 5 ) = 0
(4.5b)
cos(5α 1 ) − cos(5α 2 ) + cos(5α 3 ) − cos(5α 4 ) + cos(5α 5 ) = 0
(4.5c)
cos(7α 1 ) − cos( 7α 2 ) + cos( 7α 3 ) − cos(7α 4 ) + cos( 7α 5 ) = 0
(4.5d)
cos(9α 1 ) − cos(9α 2 ) + cos(9α 3 ) − cos(9α 4 ) + cos(9α 5 ) = 0
(4.5e)
cos(α 1 ) − cos(α 2 ) + cos(α 3 ) − cos(α 4 ) + cos(α 5 ) = 0.85
To solve these switching angles, the Newton-Raphson method in 3.2.3.1 is
applied, and the following matrices are implemented:
66
1) The switching angle matrix,
α j = [α 1 j , α 2 j , α 3 j , α 4 j , α 5 j ]T
(4.6a)
2) The nonlinear system matrix,
 cos(α 1 j ) − cos(α 2 j ) + cos(α 3 j ) − cos(α 4 j ) + cos(α 5 j ) 

j
j
j
j
j 
 cos(3α 1 ) − cos(3α 2 ) + cos(3α 3 ) − cos(3α 4 ) + cos(3α 5 ) 
F j =  cos(5α 1 j ) − cos(5α 2 j ) + cos(5α 3 j ) − cos(5α 4 j ) + cos(5α 5 j ) 


j
j
j
j
j
cos( 7α 1 ) − cos( 7α 2 ) + cos( 7α 3 ) − cos( 7α 4 ) + cos( 7α 5 ) 
 cos(9α j ) − cos(9α j ) + cos(9α j ) − cos(9α j ) + cos(9α j ) 
1
2
3
4
5


(4.6b)
and
 − sin(α 1 j )

j
j
 − 3 sin( 3α 1 )
 ∂f 
j

 ∂α  =  − 5 sin( 5α 1 )
 
j
− 7 sin( 7α 1 )
 − 9 sin( 9α j )
1

+ sin(α 2 )
− sin(α 3 )
j
j
+ 3 sin(3α 2 ) − 3 sin( 3α 3 )
j
j
+ 5 sin(5α 2 ) − 5 sin( 5α 3 )
j
j
+ 7 sin( 7α 2 ) − 7 sin( 7α 3 )
j
j
+ 9 sin(9α 2 ) − 9 sin( 9α 3 )
j
j
+ sin(α 4 )
j
+ 3 sin( 3α 4 )
j
+ 5 sin( 5α 4 )
j
+ 7 sin( 7α 4 )
j
+ 9 sin( 9α 4 )
j
j
− sin(α 5 ) 
j 
− 3 sin( 3α 5 ) 
j
− 5 sin( 5α 5 ) 

j
− 7 sin( 7α 5 )
j
− 9 sin( 9α 5 ) 
(4.6c)
3) The corresponding harmonic amplitude matrix,
 (0.85)π
T =
 4

0 0 0 0

T
(4.6d)
Thus, equation (4.6a) to (4.6d) can be rewritten in the following matrix format:
F (α ) = T
(4.7)
By using matrices (4.6a) to (4.6d) and the Newton-Raphson method, the statement
of algorithm is shown as follows:
1) Guess a set of initial values for α j with j = 0
Assume
α 0 = [α 1 , α 2 , α 3 , α 4 , α 5 ]T
(4.8)
F (α 0 ) = F 0
(4.9)
0
0
0
0
0
2) Calculate the value of
3) Linearize equation (4.7) about α 0
67
0
 ∂f 
F +   dα 0 = T
 ∂α 
0
[
dα 0 = dα 1 0
and
dα 2 0
dα 3 0
dα 4 0
(4.10)
dα 5 0
]
T
(4.11)
4) Solve dα 0 from equation (4.10), i.e.
0
 ∂f 
dα = INV   (T − F 0 )
 ∂α 
0
0
 ∂f 
 ∂f 
where INV   is the inverse matrix of  
 ∂α 
 ∂α 
(4.12)
0
5) As updated the initial values,
α
j +1
= α j + dα
j
(4.13)
6) Repeat the process for equations (4.9) to (4.13), until dα j is satisfied to the desired
degree of accuracy, and the solutions must satisfy the condition:
α1 < α 2 < α 3 < α 4 < α 5 <
π
2
(4.14)
By using MATLAB program, the results of iteration are tabulated in Table 4.1.
From Table 4.1, the final solutions, which are switching angles in degree, are
α 1 = 22.5835
α 2 = 33.6015
α 3 = 46.6433
α 4 = 68.4980
α 5 = 75.0978
68
( 4.15)
Table 4.1 The results from MATLAB program.
Iteration index
j=0
j=1
j=2
j=3
j=4
j=5
α
j
F
(degree)
20.0000
30.0000
50.0000
70.0000
80.0000
23.0612
34.0541
46.6137
69.3323
76.0393
22.5851
33.6040
46.6568
68.5420
75.1310
22.5835
33.6016
46.6434
68.4980
75.0978
22.5835
33.6015
46.6433
68.4980
75.0978
22.5835
33.6015
46.6433
68.4980
75.0978
j
df
0.5481
0.0000
0.1316
0.7879
0.0000
0.6668
0.0172
-0.0772
-0.0044
-0.0567
0.6676
-0.0002
-0.0011
0.0043
-0.0052
0.6676
0.0000
0.0000
0.0000
0.0000
0.6676
0.0000
0.0000
0.0000
0.0000
j
3.0612
4.0541
-3.3863
-0.6677
-3.9607
-0.4761
-0.4500
0.0431
-0.7903
-0.9083
-0.0016
-0.0024
-0.0134
-0.0440
-0.0332
-0.00002
-0.00005
-0.00006
-0.00001
0.00001
-0.0939e-9
-0.1345e-9
-0.0010e-9
0.1000e-9
0.0518e-9
By applying these switching angles, the 3rd, 5th, 7th, and 9th harmonics will be
eliminated from the output voltage waveform. The output voltage THD, which is also
calculated by MATLAB, is 65.15%. The results will be simulated and verified by
PSPICE, which will be shown in following section.
69
4.2.4 Verify the results with PSPICE
A three-level five-switching angle SHE PWM output waveform is synthesized by
an H-bridge voltage source inverter. The calculated switching angle in example 4.1 are
employed in the simulation. The H-bridge inverter is supplied by a 400V dc source and
operates at 400Hz. The given modulation index is 0.85, and the expected fundamental
component of the output voltage is 340V. After running PSPICE, the output voltage
waveform and its frequency spectrum are shown in Fig. 4.4 and 4.5, respectively.
From the spectrum in Fig. 4.5, the 3rd, 5th, 7th, and 9th harmonic are eliminated
from the output voltage waveform as expected. The amplitude of the fundamental is
340V, which is 85 percent of the dc source value. The first surplus harmonic is the 11th,
whose amplitude is approximately 155V or 46 percent of the fundament component. The
voltage THD, which is equal to the THD from MATLAB program, is shown in the output
file of PSPICE in Fig. 4.6.
Figure 4.4 The output voltage waveform.
70
Figure 4.5 The frequency spectrum of the output waveform shown in Fig 4.4.
**** 08/03/99 10:48:17 ********* NT Evaluation PSpice (July 1997)
* C:\My Documents\Thesis\she5angle_M85_sph.sch
****
FOURIER ANALYSIS
TEMPERATURE =
27.000 DEG C
*****************************************************************************
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(VoutA)
DC COMPONENT =
HARMONIC
NO
0.000000E+00
FREQUENCY
(HZ)
FOURIER
COMPONENT
NORMALIZED
COMPONENT
PHASE
(DEG)
NORMALIZED
PHASE (DEG)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
4.000E+02
8.000E+02
1.200E+03
1.600E+03
2.000E+03
2.400E+03
2.800E+03
3.200E+03
3.600E+03
4.000E+03
4.400E+03
4.800E+03
5.200E+03
5.600E+03
6.000E+03
3.400E+02
1.465E-02
2.411E-02
6.953E-03
1.240E-02
1.819E-02
5.541E-02
2.263E-02
2.095E-02
4.210E-03
1.554E+02
3.015E-02
2.029E+01
2.285E-02
9.364E+01
1.000E+00
4.310E-05
7.091E-05
2.045E-05
3.648E-05
5.350E-05
1.630E-04
6.656E-05
6.160E-05
1.238E-05
4.571E-01
8.868E-05
5.967E-02
6.722E-05
2.754E-01
-7.192E-02
6.933E+01
-1.660E+02
-8.747E+01
-4.176E+01
-1.292E+02
5.520E-01
1.238E+02
4.673E+01
6.963E+01
1.792E+02
5.292E+01
-9.411E-01
-3.565E+01
-1.095E+00
0.000E+00
6.940E+01
-1.659E+02
-8.740E+01
-4.169E+01
-1.291E+02
6.240E-01
1.239E+02
4.680E+01
6.970E+01
1.793E+02
5.299E+01
-8.692E-01
-3.558E+01
-1.023E+00
62
63
2.480E+04
2.520E+04
2.109E-02
3.083E+00
6.202E-05
9.068E-03
-1.122E+02
-4.545E+00
-1.122E+02
-4.473E+00
M
TOTAL HARMONIC DISTORTION =
6.513426E+01 PERCENT
Figure 4.6 The harmonic components and THD from PSPICE output file.
71
4.3
Conclusions
The selective harmonic eliminated PWM technique is introduced. The output
waveform is shopped N times per quarter-cycle. The N-1 harmonics can be removed
from the waveform and the fundamental amplitude can be controlled.
The simulated results agree with the calculated results very well.
As discussed in Chapter 3, the condition that the solutions of OHSW must be
satisfied
is
α 1 , α 2 , α 3 ,α 4 <
α1 < α 2 < α 3 < α 4 < α 5 <
π
.
2
Compared
to
the
SHE
PWM
condition,
π
, it is easier to get the solutions because its condition has more
2
degree of freedom. This is an advantage of OHSW over SHE PWM technique.
72